Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance

Where, d = distance
A = amplitude
Put the value into the formula


Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Answer:
A. False, frequency can increase or decrease wavelength.
For example: a high frequency would mean there are shorter wavelengths that occur in a period. Meanwhile, a low frequency would indicate that the wavelengths are longer and in longer periods.
Answer:
x = 2
Explanation:
if it was -7 = the square root of both 2x-9 together, it would be false.
if it was square root of just 2x in the equation, the answer is:
x = 2
°°°°°°°°°
-7 = √2x - 9
-√2x = -9 + 7
√-2x = -2
√2x = 2
2x = 4
x = 2